P(Y=0)=\sum_{k=0}^{\infty}P(X=\frac{1}{2k+1})=3\sum_{k=0}^{\infty}({\frac{1}{4}})^{2k+1}=
3\frac{1}{4}\sum_{k=0}^{\infty}(\frac{1}{4})^{2k}=3\frac{1}{4}\sum_{k=0}^{\infty}(\frac{1}{16})^{k}=
\frac{3}{4}\frac{1}{1-\frac{1}{16}}=\frac{3}{4}\frac{16}{15}=\frac{4}{5}